Calculate [H3O+] for each solution.
pH = 8.25
pH = 11.63
pH = 2.37
pH = 1.32
PH = -log[H3O+]
8.25 = -log[H3O+]
[H3O+] = 10-8.25
[H3O+] = 5.62*10-9 M
pH = 11.63
PH = -log[H3O+]
11.63 = -log[H3O+]
[H3O+] = 10-11.63
[H3O+] = 2.344*10-12 M
pH = 2.37
PH = -log[H3O+]
2.37 = -log[H3O+]
[H3O+] = 10-2.37
[H3O+] = 4.26*10-3 M
pH = 1.32
PH = -log[H3O+]
1.32 = -log[H3O+]
[H3O+] = 10-1.32
[H3O+] = 4.78*10-2 M
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